How many grams of sodium nitrate are needed to prepare a 0.965 M sodium nitrate solution that contains 1,640 g of water?

134.5 g

Explanation:

Let us recall that 1g of water = 1cm3 of water

Hence;

1,640 g of water =  1,640 cm3 of water or 1.64 L

This is now the volume of the solution

Now;

m/M = CV

m = mass of the solute

M = molar mass of the solute

C = concentration of the solution

V = volume of the solution

m/84.9947 g/mol = 0.965 M *  1.64 L

m = 0.965 M *  1.64 L * 84.9947 g/mol

m = 134.5 g